80. Find all values of c that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.
c. f(x) = x³/ (3 - 4x), g(x) = x², (a, b) = (0, 3)

Verified step by step guidance

1. Recall Cauchy's Mean Value Theorem (CMVT): If functions \(f\) and \(g\) are continuous on \([a,b]\) and differentiable on \((a,b)\), and \(g'(x) \neq 0\) on \((a,b)\), then there exists at least one \(c \in (a,b)\) such that \[\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}.\]
2. Verify that \(f(x) = \frac{x^3}{3 - 4x}\) and \(g(x) = x^2\) are continuous on \([0,3]\) and differentiable on \((0,3)\). Note any points where the denominator \(3 - 4x\) might be zero and exclude those if necessary.
3. Calculate the values \(f(a)\), \(f(b)\), \(g(a)\), and \(g(b)\) for \(a=0\) and \(b=3\): \[f(0) = \frac{0^3}{3 - 4 \cdot 0} = 0,\] \[f(3) = \frac{3^3}{3 - 4 \cdot 3} = \frac{27}{3 - 12} = \frac{27}{-9},\] \[g(0) = 0^2 = 0,\] \[g(3) = 3^2 = 9.\]
4. Find the derivatives \(f'(x)\) and \(g'(x)\): For \(f(x) = \frac{x^3}{3 - 4x}\), use the quotient rule: \[f'(x) = \frac{(3x^2)(3 - 4x) - x^3(-4)}{(3 - 4x)^2} = \frac{3x^2(3 - 4x) + 4x^3}{(3 - 4x)^2}.\] For \(g(x) = x^2\), the derivative is: \[g'(x) = 2x.\]
5. Set up the equation from CMVT: \[\frac{f'(c)}{g'(c)} = \frac{f(3) - f(0)}{g(3) - g(0)} = \frac{\frac{27}{-9} - 0}{9 - 0} = \frac{-3}{9} = -\frac{1}{3}.\] Substitute the derivatives: \[\frac{\frac{3c^2(3 - 4c) + 4c^3}{(3 - 4c)^2}}{2c} = -\frac{1}{3}.\] Simplify and solve this equation for \(c\) in the interval \((0,3)\), excluding any points where the denominator is zero.

Key Concepts

Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem generalizes the Mean Value Theorem by relating two functions. It states that if functions f and g are continuous on [a, b] and differentiable on (a, b), and g'(x) ≠ 0 on (a, b), then there exists at least one c in (a, b) such that (f(b)-f(a))/(g(b)-g(a)) = f'(c)/g'(c).

Differentiability and Continuity of Functions

To apply Cauchy's Mean Value Theorem, both functions must be continuous on the closed interval and differentiable on the open interval. This ensures the existence of derivatives and the validity of the theorem. Checking these conditions for f(x) = x³/(3 - 4x) and g(x) = x² on (0, 3) is essential.

Finding and Solving for c

After confirming the theorem's conditions, compute f'(x) and g'(x), then set the ratio of their differences equal to the ratio of their derivatives at c. This leads to an equation in c, which must be solved within the interval (0, 3) to find all valid values satisfying the theorem.